How do you find the important points to graph #f(x)=2x^2#?

1 Answer
Tony B
Jul 10, 2018

See explanation

Explanation:

Set #y=f(x)=2x^2#

Compare to the standardised form: #y=ax^2+bx+c#
This equation #color(white)("dddddfdd") ->color(white)("ddddd")y=2x^2+0x+0 #

Key points:

#a>0 -> ("positive")# so the graph is of form #uu#

#c-># y-intercept #=0#

#x_("vertex") -> (-1/2)xx b/acolor(white)("ddd") =color(white)("ddd") (-1/2)xx0/2color(white)("d")=color(white)("d")0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As #x_("vertex")=0# the axis of symmetry is the y-axis

As #a>0# then the general shape is #uu# with the y-axis in the middle.

Couple this with #c=0# and it means that the vertex is at #(x,y)=(0.0)#

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